为 A xor X = B + X 找到解决方案的算法

Question

Given integer A and B, find integer X so that:

A,B < 2*1e18

A xor X = B + X

我非常怀疑能否用数学来求解这个方程。这是我 3 年前遇到的一个编码问题，即使现在我自己也无法解决。

到目前为止我的代码：（这是蛮力解决方案）

#include <iostream>

using namespace std;

int main()
{

    unsigned long long a, b;
    cin >> a >> b;
    for (unsigned long long x = 1; x < max(a, b); x++) {
        unsigned long long c = a ^ x;
        unsigned long long d = b + x;
        if (c == d) {
            cout << x << endl;
            break;
            return 0;
        }
    }

    cout << -1; //if no such integer exists

    return 0;
}

Answer 1

不是很难，你只需要从小处着想：假设我们用二进制写A、B和X，Aᵢ是对应的值到最右边的 2ⁱ 位。

我们知道：Aₒ ⊕ Xₒ = Bₒ + Xₒ.

让我们用一个例子来了解如何计算：A = 15 和 B = 6。转换为二进制：

A = 1 1 1 1           B = 0 1 1 0
X = a b c d           X = a b c d

现在我们有了一些可能性。下面分析一下A和B最右边的位：

1 ⊕ d = 0 + d

我们知道d只能是0或1，所以：

for d = 0
1 ⊕ d = 0 + d    =>    1 ⊕ 0 = 0 + 0    =>    1 = 0 (not possible)

for d = 1
1 ⊕ d = 0 + d    =>    1 ⊕ 1 = 0 + 1    =>    0 = 1 (not possible)

值得注意的是，XOR 的行为就像二进制和（区别在于 XOR 不会为下一个位和创建结转）：

    XOR           SUM
0 ⊕ 0 = 0  |   0 + 0 = 0
0 ⊕ 1 = 1  |   0 + 1 = 1
1 ⊕ 0 = 1  |   1 + 0 = 1
1 ⊕ 1 = 0  |   1 + 1 = 0

所以并不总能找到满足 A ⊕ X = B + X 的 X，因为不存在满足 1 + d = 0 + d.

的值 d

总之，如果X存在，就这样找吧，从右到左，一点一点找。

完整示例

A = 15, B = 7:

A = 1 1 1 1           B = 0 1 1 1
X = a b c d           X = a b c d

1 ⊕ d = 1 + d

这里，d = 0和d = 1都适用，那又如何呢？我们需要检查下一位。假设 d = 1:

A = 1 1 1 1           B = 0 1 1 1
X = a b c d           X = a b c d

1 ⊕ d = 1 + d    =>    1 ⊕ 1 = 1 + 1    =>    0 = 0 (possible)

BUT 1 + 1 = 0 generates a carryover for the next bit sum:

Instead of 1 ⊕ c = 1 + c, we have 1 ⊕ c = 1 + c (+1) =
                                   1 ⊕ c = c  (not possible)

所以在这种情况下，d 必须为 0。

carryover                              0
         A = 1 1 1 1           B = 0 1 1 1
         X = a b 0 0           X = a b 0 0
        -----------------------------------
                   0                     0

we know that c must be 0:

carryover                            0 0
         A = 1 1 1 1           B = 0 1 1 1
         X = a b 0 0           X = a b 0 0
        -----------------------------------
                 1 1                   1 1

但是b呢？我们需要像往常一样检查下一位：

if b = 0, there won't be a carryover, so we'll have:

1 ⊕ a = 0 + a  (and this is not possible)

so we try b = 1:

1 ⊕ b = 1 + b    =>    1 ⊕ 1 = 1 + 1    =>    0 = 0 (with carryover)

现在，a：

carryover                          1 0 0
         A = 1 1 1 1           B = 0 1 1 1
         X = a 1 0 0           X = a 1 0 0
        -----------------------------------
               0 0 0                 0 0 0


1 ⊕ a = 0 + a (+1)    =>    1 ⊕ a = 1 + a

这里a可以是0和1，但必须是0，以免在和B + X.

中出现结转

然后，X = 0 1 0 0，因此 X = 4。

代码

#include <iostream>
using namespace std;

inline int bit(int a, int n) {
    if(n > 31) return 0; 
    return (a & ( 1 << n )) >> n; 
}

int main(){
    int A = 19;
    int B = 7;

    int X = 0;
    int carryover = 0;
    int aCurrent, aNext, bCurrent, bNext;

    for(int i = 0; i < 32; i++){
        aCurrent =  bit(A, i);      bCurrent =  bit(B, i);
        aNext =     bit(A, i + 1);  bNext =     bit(B, i + 1);

        if(aCurrent == 0 && bCurrent == 0){
            if(carryover) {X = -1; break;}
            if(aNext != bNext){
                X += 1 << i;
            }
            carryover = 0;
        }
        else if(aCurrent == 0 && bCurrent == 1){
            if(!carryover) {X = -1; break;}
            if(aNext == bNext){
                X += 1 << i;
            }
            carryover = 1;
        }
        else if(aCurrent == 1 && bCurrent == 0){
            if(!carryover) {X = -1; break;}
            if(aNext != bNext){
                X += 1 << i;
                carryover = 1;
            }
            else {
                carryover = 0;
            }
        }
        else if(aCurrent == 1 && bCurrent == 1){
            if(carryover) {X = -1; break;}
            if(aNext != bNext){
                X += 1 << i;
                carryover = 1;
            }
            else {
                carryover = 0;
            }
        }

    }

    if(X != -1) cout<<"X = "<<X<<endl;
    else cout<<"X doesnt exist"<<endl;

    return 0;
}

你可以测试一下here。

Answer 2

注意A + X == (A xor X) + ((A and X)<<1)。所以：

A xor X = A + X - ((A and X)<<1) = B + X
A - B = (A and X)<<1

我们有：

(A - B) and not (A<<1) = 0    (All bits in (A - B) are also set in (A<<1))
(A - B)>>1 = A and X

如果满足条件，对于任何没有在 A 中设置位的整数 Y，(((A - B)>>1) 或 Y) 是一个解。如果你只想要一个解决方案，你可以使用 ((A - B)>>1)，其中 Y = 0。否则没有解决方案。

int solve(int a, int b){
    int x = (a - b) >> 1;
    if ((a ^ x) == b + x)
        return x;
    else
        return ERROR;
}

为 A xor X = B + X 找到解决方案的算法

Algorithm to find a solution for A xor X = B + X

c++

xor